This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Taught By

Dr. Bonnie H. Ferri

Professor

Dr. Joyelle Harris

Director of the Engineering for Social Innovation Center

Dr Mary Ann Weitnauer

Transcript

The topic of this problem is Thevenin's Analysis. And we're going to work with circuits with independent sources. The problem is to use Thevenin's equivalent circuit to determine Vo in the circuit shown below. The Vo is the voltage, the output voltage across the 8 kilo ohm load resistance. We have two independent sources, a 12 V source and a 2 mA source in the problem. So, the first thing that we need to know is that Thevenin's equivalent circuit, Has this form. It has an open circuit voltage which is part of the equivalent circuit, it has a Thevenin's equivalent resistance which is part of the equivalent circuit. And then we have a load resistance or a load circuit, in our case, we're going to reduce it to just a single resistance on the right hand side of the circuit. So, we have to come up with a circuit which looks like this, with Voc and R Th in order to put it in the form of a Thevenin's equivalent circuit so that we can find the voltage drop, in our case, across our load. So once we have it in this form, it's going to be simply a voltage division between R sub l and R Th to find Vo. The trick is to get it in this form. So we want to first find Voc, be the first thing that we look for. You can also look for R Th first, it's just what I've chosen to do. So we had to redraw our circuit for the open circuit condition. So we'll have our 12 V source and we have all the other components of our circuit in it as well, except for the 8K resistor, we've taken the load and we've open circuited the load. So we have a Voc that we're measuring. We still have our 2 mA source and our 6 kilo ohm resistor in the center. And so this is what our equivalent circuit for finding Voc looks like. Labeling all the resistors and all the voltage sources and current sources, this is what we get looking for Voc. So we recognize first of all that there is no current flowing in the 4 kilo ohm resistor, it's an open circuit condition. So there is no voltage drop associated with this 4 kilo ohm resistor. So the voltage drop that we're looking for is the voltage drop from this point of the circuit to the ground point in the circuit, and so that's what we're going to try to find is that particular voltage. To find that voltage we have a number of different ways to do it. We can use mesh analysis, we can use nodal analysis, it's really up to us what we end up doing there. So if we want to, we can look at solving the circuit using mesh analysis and look for voltage at this point, and summing, perhaps, summing the currents into that node. So we have a current flowing through the 3K resistor that we summed in, with this current and with this current. So let's do that. Starting with the current through the 3K resistor is going to be 12 V, which is a voltage here, minus, I'll call that node 1, it's going to be 12 V, minus V sub 1. Divided by 3 kilo ohm. The current up through the 6K resistor is going to be zero, which is a voltage at the bottom of the circuit, it's a ground node, minus V1 Divided by 6k. And we have 2 mA following in here, and so this is going to be + 2 mA = 0. So we can find our V1 very easily using this approach. And once we find V1, we now that Voc, Is going to be equal to what? It's going to be equal to the voltage V1 times 6, which is going to be the voltage V1 plus the voltage drop across the 2 kilo ohm resistor. Because this voltage Voc is again, the same voltage from this point to this point which is the voltage drop across the 2 kilo ohm resistor plus the voltage drop across the 6 kilo ohm resistor which we know is already V1. So Voc is equal to V1 plus the voltage drop here which is 2 mA times 2 kilo ohms. So we solve for V1 in our first equation that we have. We plug it into our Voc equation and what we end up with is a Voc = 16 V. Once we have that 16 V, we can then use that in our Thevenin's equivalent circuit. But before we are finished, we have to also find R Th. To find R Th, we take our original circuit, we take the load out of the circuit and we look for the equivalent resistance looking back in to the remaining part of the circuit. To find that equivalent resistance, we short circuit all the voltage sources, and we'll open circuit all the current sources. And again, we've taken the load which is 8 kilo ohms out of our circuit. So, R Th is going to be the equivalent resistance, looking back into our circuit, with the current source as open circuited, the voltage source is short circuited. So, this is our equivalent circuit for finding R Th. So R Th is going to be 3 kilo ohms in parallel with 6 kilo ohms + 2 kilo ohms + 4 kilo ohms. So a 3 in parallel with 6 is 2, plus 2 is 4, plus 4 is 8 kilo ohms for R Th. So now if we have R Th equal to 8 Kilo ohms and a Voc equal to 16 V, we can redraw our Thevenin's equivalent circuit putting those values in for our sources and our resistances. And putting our load back in as well. And now again we're looking for Vo in this circuit and Vo is a voltage drop across the 8K resistor. We have two 8K resistors and a voltage source which is 16 V. So we know that Vo is going to be half of what the source is, because half of it is dropped across one 8K resistor, and the other is across the other 8K resistor. We can use voltage division to confirm that. And it's going to be the initial voltage times the resistor that we're looking for, the voltage drop across divided by the sum of the two resistances. That's a simple voltage division. So, we get a Vo which is equal to 8 V.

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